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Some silly modeling question 
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#1
Feb2609, 11:00 PM

P: 53

so my bro ahd an algebra 2 midterm and a question was this:
) A printer costs $35,000 (how old are these questions lol) but it depreciates 5% a year. What is the value by the 4th year? ok so being an overtly complex person i tried to model a differential equation of this just for the kicks. At first I thought dP/dt=0.05P where P is price and the solution is Po*exp(0.5t) where Po=35000 however this is wrong. So I assumed I did not know r from dP/dt=rP and worked the problem by finding the initial values. so the solution gives me P=exp(.051293t)Po which is correct. I dont grasp intuitively the answer. why is dP/dt=.051293P when the problem says it goes down 0.5 each year so I assume P changes over time by 0.05P per year? 


#2
Feb2609, 11:32 PM

P: 287

The correct equation is, of course.
P = P_0 * (0.95)^t. Another way of writing this is: P = P_0 * exp(ln(0.95)*t). As you probably have already guessed, ln(0.95) is the number you need. ln(0.95) = .051293, approximately. 


#3
Feb2609, 11:36 PM

P: 53

yeah, thats true. i already knew that but, why isnt dP/dt=0.5P!!! it does not make sense in my mind. it seems that the changing rate of P is 0.5 P per year



#4
Feb2709, 07:30 AM

P: 287

Some silly modeling question
Almost certainly because 0.5 isn't the instantaneous rate, but the "average" yearly rate.
What you should say is that the integral from 0 to 1 of e^r equals 0.05... then e^r = 0.95 and r = ln(0.95). In fact, that's exactly it. You can't substitute the average (or cumulative, to put it in better terms) rate for the instantaneous rate. 


#5
Feb2709, 01:51 PM

P: 53

that makes a lot of sense! because i can still integrate from 0 to .1 and that wouldnt obviously be right.



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